Question

The speeds of overtaking of overtakten vehicle are
80 and 60 kmph respectively if acceleration of
overtaking vehicle is 2.5 kmph per second,
Calculate the safe passing sight Distance for
a .one way traffic
b.two way traffic​

Answer 1

Answer:-

$\red{\bigstar}$ Overtaking Sight Distance for

1. One-way traffic $\large\leadsto\boxed{\rm\purple{248.68 \: m}}$

2. Two-way traffic $\large\leadsto\boxed{\rm\purple{464.02 \: m}}$

• Given:-

$\sf V = 80 kmph$

→ $\sf V = 80 \times \dfrac{5}{18}$

→ V = 22.23 m/s

$\sf V_b = 60 kmph$

→ $\sf V_b = 60 \times \dfrac{5}{18}$

→ $\bf V_b = 16.67 m/s$

$\sf a = 2.5 kmph$

→ $\sf a = 2.5 \times \dfrac{5}{18}$

→ $\bf a = 0.695 m/s$

• To Find:-

Overtaking Sight Distance [O.S.D] for

1. one-way traffic

2. two-way traffic

• Solution:-

1. One-way Traffic:-

We know, O.S.D for one way traffic:-

$\large\boxed{\red{\bf O.S.D = d_1 + d_2}}$

where,

$\boxed{\green{\bf d_1 = V_b \times t}}$

→ $\sf d_1 = 16.67 \times 2$

→ $\sf d_1 = 33.34$

★ Time of reaction[t] is assumed as 2 sec. in O.S.D

and

$\boxed{\green{\bf d_2 = V_b \times T + 2S}}$

Now, we know

$\boxed{\green{\bf S = (0.7 V_b + 6)}}$

→ $\sf S = 0.7 \times 16.67 + 6$

→ $\sf S = 11.669 + 6$

→ $\bf S = 17.669 m$

Also,

$\boxed{\green{\bf T = \sqrt{\dfrac{4S}{a}}}}$

→ $\sf T = \sqrt{\dfrac{4 \times 17.669}{0.695}}$

→ $\sf T = \sqrt{\dfrac{70.676}{0.695}}$

→ $\sf T = \sqrt{101.69}$

→ $\bf T = 10.08 sec.$

Substituting in the values in d2:-

→ $\sf d_2 = 16.67 \times 10.8 + 2\times 17.67$

→ $\sf d_2 = 180 + 35.34$

→ $\sf d_2 = 215.34$

Hence,

O.S.D = $\sf d_1 + d_2$

→ 33.34 + 215.34 m

✴ 248.68 m

2. Two-way traffic:-

We know, O.S.D for two-way traffic:-

$\large\boxed{\red{\bf O.S.D = d_1 + d_2 + d_3}}$

We already know the value of $\bf d_1 \: and \: d_2$

Now,

$\boxed{\green{\bf d_3 = V \times T}}$

→ $\sf 22.23 \times 10.08$

→ $\bf 224.07$

Hence,

$\bf O.S.D = d_1 + d_2 + d_3$

→ $\sf O.S.D = 33.34 + 215.34 + 224.07$

→ $\sf 248.68 + 215.34$

✴ 464.02 m